Forcing a Particle to Rest: Analyzing Position-Dependent Force

[JennV]

Homework Statement

A 30.0 kg particle is initially at x=0 and has a velocity of 5.50 m/s. It encounters a position-dependent net force F(x) described by the graph shown (the graph continues indefinitely in the manner shown past x=20.0 m).

Diagram:

http://img137.imageshack.us/img137/9502/forcegraph.jpg

A.) How fast is the particle moving at x = 3.00 m?
I have already obtained an answer of v = 5.95m/s and it is correct

This is what I would like help for:
B.) How far does the particle travel along the x-axis before being brought to rest by the force? (Note: The answer is not 8 m or 20 m. You may safely assume for simplicity that the final position is greater than 8 m.)

Homework Equations

Wtotal=0.5mvf^2 - 0.5mvi^2
W=F*deltaX

The Attempt at a Solution

Wtotal = 0.5(30)(0)^2 - 0.5(30)(5.95)^2 = -453.75J
So does xf=Wtotal/slope of graph (F) ?
xf=-453.75J/-4 ?
Am I doing this correct?

THANKS!

 

[The legend]

Wtotal = 0.5(30)(0)^2 - 0.5(30)(5.95)^2 = -453.75J

Recheck the bold part...(vi is not equal to 5.95).

So does xf=Wtotal/slope of graph (F) ?

Yes, I think.

 

[PhanthomJay]

I am unsure if you used calculus to solve for part a correctly, or whether you used an average force over that displacement interval, or whether you used the area under the graph between those points to solve for the total work. Anyway, proceed in the same manner to solve for the displacement at v = 0, by solving for the work done.